#1

Can someone help me with this?

1) |a+b| </= (Lesser than or equal to) |a| + |b|

Prove

I tried for some time, but I cant get this... :S

For the next few I have to find the solution sets for x

2) 4 - |x| > |2x|

3) |x/3 - 1| > |x|

The concept of the modulus signs is new to me, and hence I'd appreciate if someone could do the above sums as example sums so I could complete the homework on my own as well as prepare for the test on monday

Thanks

1) |a+b| </= (Lesser than or equal to) |a| + |b|

Prove

I tried for some time, but I cant get this... :S

For the next few I have to find the solution sets for x

2) 4 - |x| > |2x|

3) |x/3 - 1| > |x|

The concept of the modulus signs is new to me, and hence I'd appreciate if someone could do the above sums as example sums so I could complete the homework on my own as well as prepare for the test on monday

Thanks

#2

lolz... i haven't covered modulus yet.

we start when we return to school in 7 weeks.

we start when we return to school in 7 weeks.

#3

Looking for something like this?

#4

Yup, but I got +4/3 -4/3 >.<

I got where I made the mistake.

Could anyone help me with the others please?

Especially 1) |a+b| < Or = |a| + |b|

Prove?

Thanks

I got where I made the mistake.

Could anyone help me with the others please?

Especially 1) |a+b| < Or = |a| + |b|

Prove?

Thanks

#5

edit, mistake at late one, it's x > 4/3

#6

Delivering 3, sorry for the messy writing

trying 1

trying 1

#7

Nice

I was getting really worked up about the third one because in 3) For case 1 where we take x to be +ve, i got a -ve answer, and vice versa, I thought I was doing it wrong

I just did the whole working and wrote not possible

I was getting really worked up about the third one because in 3) For case 1 where we take x to be +ve, i got a -ve answer, and vice versa, I thought I was doing it wrong

I just did the whole working and wrote not possible

#8

No idea about 1), I can only do it when I give them a number :/

#9

Aah, okay.

Thanks A LOT though

Could you pm me if you DO get it?

Thanks A LOT though

Could you pm me if you DO get it?

#10

It's in my book of 2 years ago but had to hand it in

#11

|a+b| < = |a| + |b|

it's called the triangle inequality |a| and |b| are the two adjacent sides of the triangle |a+b| is the third side. Logically, |a+b| is bigger than |a| and |b| but is either < or = to |a|+|b|

it's called the triangle inequality |a| and |b| are the two adjacent sides of the triangle |a+b| is the third side. Logically, |a+b| is bigger than |a| and |b| but is either < or = to |a|+|b|

#12

@Anonymoos

Its okay lol :P

@rock.freak667

I'll try that, thanks

Its okay lol :P

@rock.freak667

I'll try that, thanks